Given, $f(x)=a \cdot \log |x|+b x^2+x$
$$
\begin{aligned}
& f^{\prime}(x)=a \cdot \frac{1}{|x|} \cdot \frac{x}{|x|}+b(2 x)+1 \\
& f^{\prime}(x)=\frac{a x}{|x|^2}+2 b x+1
\end{aligned}
$$
Given that $x=-1$ is one of extremity of $f(x)$.
$$
\begin{aligned}
\Rightarrow f^{\prime}(-1)= & 0 \\
& \frac{a(-1)}{1}+2 b(-1)+1=0
\end{aligned}
$$


Also, given $x=2$ is the other extremity of $f(x)$.
$$
\begin{aligned}
f^{\prime}(2) & =0 \\
\frac{a(2)}{4}+2 b(2)+1 & =0 \\
\frac{a}{2}+4 b+1 & =0
\end{aligned}
$$

Adding on Eqs. (i) and (ii), we get
$$
\begin{array}{rlrl}
\Rightarrow & & 6 b+3 & =0 \\
\Rightarrow & b & =-\frac{1}{2}
\end{array}
$$

Substituting $b=-\frac{1}{2}$ in Eq. (i), we get
$$
\begin{aligned}
a & =2 \\
\therefore(a, b) & =\left(2,-\frac{1}{2}\right)
\end{aligned}
$$
$\therefore$ Hence option (b) is correct.